Catchup and Pass Example Problem

Solution:

We know that Sarah is running at a speed of $800\frac{\text{feet}}{\text{minute}}$ and Joon is running at a speed of $\mathrm{1,200}\frac{\text{feet}}{\text{minute}}$.

The amount of time Joon runs is equal to the amount of time Sarah runs. We can let t = this time. If you are wondering why the time is t for both Sara and Joon, please review the note at the bottom of the solution.

Now that we have a rate and time for Joon and Sarah, we can determine their distances using the distance formula:

$\begin{array}{l}\Rightarrow {\text{distance}}_{\mathrm{J}\mathrm{o}\mathrm{o}\mathrm{n}}=\mathrm{1,200}\frac{\text{feet}}{\text{minute}}\times \text{t minutes}=\mathrm{1,200}\mathrm{t}\text{ feet}\\ \Rightarrow {\text{distance}}_{\mathrm{S}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{h}}=800\frac{\text{feet}}{\text{minute}}\times \text{t minutes}=800\mathrm{t}\text{ feet}\end{array}$

In a normal catch-up problem we would set the two distances equal and determine t. However, in this problem, Joon and Sarah do not cover the same distance. In fact, 
Joon actually runs 120 more feet than Sarah because Joon starts 80 feet behind Sarah and then passes her by 40 feet. In other words, not only must Joon run the distance that Sarah runs, but she must also run the additional 120 feet (40 feet just to catch Sarah and then 80 more feet than what Sarah runs). 

If this is confusing, just remember, in this particular problem the equation must be:

Person Traveling the Lesser Distance + 120 = Person Traveling the Greater Distance

Since Sarah traveled the lesser distance, and since Joon traveled the greater distance, we have:

$\begin{array}{l}\Rightarrow {\text{distance}}_{\mathrm{J}\mathrm{o}\mathrm{o}\mathrm{n}}={\text{distance}}_{\mathrm{S}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{h}}+120\text{ feet}\\ \Rightarrow \mathrm{1,200}\mathrm{t}=800\mathrm{t}+120\\ \Rightarrow 400\mathrm{t}=120\\ \Rightarrow \mathrm{t}=\frac{120}{400}\text{=}\frac{3}{10}\text{minute}\end{array}$

We now must convert this time from minutes to seconds:

$\Rightarrow \text{Time in Seconds}=\left(\frac{\frac{3}{10}\text{ minute}}{1}\right)\times \left(\frac{\text{60 seconds}}{\text{1 minute}}\right)$

$\Rightarrow \mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\text{ in Seconds}=\frac{\frac{3}{10}\cancel{\text{minute}}\times \text{ 60 seconds}}{1\cancel{\text{minute}}}=18\text{ seconds}$

The total time it takes Joon to catch up and pass Sarah is 18 seconds.

Note: If you are wondering why the times for both Joon and Sarah are the same, we can include some more context as to exactly what is happening in this problem. Imagine that Sarah and Joon were both running around a track and, at the very moment Joon was 80 feet behind Sarah, Joon started a stopwatch. She then kept the stopwatch running until the exact moment (18 seconds later) that she was now 40 feet ahead of Sarah. So, we see that they both were running for 18 seconds while this particular “event” took place.